Redox Reactions

1
Redox Reactions

• Electrons, electrons everywhere!
• It would be so much easier of they just stayed in
  one place, wouldnt it?

2
Definitions

• Oxidation
• Originally meant the combination of any element
  with oxygen
• Now includes any reaction that involves the loss
  of electrons
• Reduction
• Originally meant the removal of oxygen from a
  metal
• Now includes any reaction that involves the gain
  of electrons

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What is a redox reaction?

• Redox a reaction that involves oxidation and
  reduction
• So what is the difference between this and the
  precipitation reactions we looked at yesterday?
• Lets look at a couple of examples

4
Example of a Non-Redox

• Lets look at a precipitation reaction
• AgCl Na2SO4 ? Ag2SO4 NaCl
• Notice that the charges of each of the ions are
  the same on both sides
• Ag1 ? Ag1
• Na1 ? Na1
• SO4-2 ? SO4-2
• Cl-1 ? Cl-1

5
Example of a Redox

• Look at this synthesis reaction
• Fe O2 ? Fe2O3
• What happens to the charge on each of the
  elements?
• Fe0 ? Fe3
• O20 ? O-2

6
Which reactions qualify as redox?

• Redox reactions reactions in which the charge of
  elements change from one side to the other
• The element that is Oxidized loses electrons,
  which means a more positive charge (the iron in
  the previous example)
• The element that is Reduced gains electrons,
  which means a more negative charge (the oxygen in
  the previous reaction)

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Assigning oxidation numbers

• Oxidation Number the apparent charge of an atom
• For ions/ionic compounds this is the same as its
  charge
• For covalent compounds this is the imaginary
  charge the atoms would have if the e- being
  shared were assigned to the atom that has the
  greater attraction for the e-
• Gold Sheet
• Lists rules for assigning oxidation numbers
• You should know these for the test

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Examples of Determining Oxidation Numbers

• What are the oxidation numbers of each element in
  KNO3?
• O -2 (rule 6)
• K 1 (rule 7)
• Since the oxidation numbers of a neutral compound
  must equal 0 (rule 8)
• (1) (N) (3-2) 0
• N 5

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Examples of Determining Oxidation Numbers
(continued)

• What are the oxidation numbers of each element in
  (NH4)1?
• H 1 (rule 4)
• Since the oxidation numbers of a polyatomic ion
  must equal its charge (rule 9)
• (N) (41) 1
• N -3

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Do these on your own

• Determine the oxidation numbers for each element
  in the following chemicals
• H2CO3
• K3PO4
• Al(NO3)3

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Answers

• H2CO3
• H1
• O-2
• C4

• K3PO4
• K1
• P5
• O-2

• Al(NO3)3
• Al3
• N5
• O-2

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Is this a redox reaction?

• Al H2O ? Al2O3 H2
• To be redox, an element must be oxidized, and an
  element must be reduced
• Assign oxidation numbers to each element to
  determine if the reaction is redox
• Al0 ?Al3
• H1?H0
• O-2? O-2
• Since the aluminum is oxidized, and the hydrogen
  is reduced, it is a redox reaction

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A couple more terms

• From the previous example, we know
• Al H2O ? Al2O3 H2
• Al0 ?Al3
• H1?H0
• The aluminum is oxidized
• The hydrogen is reduced
• H2O is called the oxidizing agent (the e-
  acceptor)
• Al is called the reducing agent (the e- donor)

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Balancing Redox Reactions

• Now that we know what they are, how do we balance
  them?
• Will it be the same as before?
• Of course not.
• Are we excited to learn more techniques that will
  improve our Chemistry knowledge?
• Of course we are.

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Where do redox reactions occur?

• Redox reactions typically take place in solutions
• Because of this redox reactions are most often
  written as net ionic equations
• If the reaction takes place in only water, you
  can balance by inspection (just like normal)
• However, if the reactions involve acidic or basic
  solutions balancing is different

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Why would a redox reaction take place in an acid
or base?

• First, lets define
• Acid H ion donator when dissolved in water
• Base H ion acceptor when dissolved in water
• When these types of solutions are present, it
  creates an opportunity for charges to get messed
  with
• Extra H ions will attract electrons
• Removing H ions creates leftover OH- ions in the
  water

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Writing Half Reactions

• In any case, when balancing a redox reaction it
  will be handy to write the reaction as two half
  reactions
• One showing only the oxidation
• One showing only the reduction
• We will treat the two half reactions separately,
  and then combine them at the end

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Writing Half Reactions Examples

• Write the half reactions for this redox
• CeCl4 Sn(NO3)2 ? SnCl4 Ce(NO3)3
• The half reactions would look like this
• Ce4 ? Ce3 (reduction)
• Sn2 ? Sn4 (oxidation)
• Notice the cancellation of the spectator ions,
  nitrate and chloride (they are net ionic eqns)

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To balance a redox reaction in an acidic solution

• Write the reaction as two half reactions
• For each half reaction
• Balance the elements except H and O
• Balance the oxygen by adding water
• Balance the hydrogen by adding H ions
• Balance the charge by adding electrons (e-)
• Multiply the half reactions to equalize the
  number of electrons in each
• Combine the half reactions and cancel anything
  identical (which should always include the
  electrons)

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To balance a redox reaction in an acidic
solution Example

• Balance the following redox reaction that occurs
  in an acidic solution
• MnO4-1 Fe2 ? Fe3 Mn2
• Write the reaction as two half reactions
• Oxidation Fe2 ? Fe3
• Reduction MnO4-1 ? Mn2

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To balance a redox reaction in an acidic
solution Example

• Fe2 ? Fe3 MnO4-1 ? Mn2
• For each half reaction
• Balance the elements except H and O
• Not needed in these half reactions
• Balance the oxygen by adding water
• MnO4-1 ? Mn2 4H2O
• Balance the hydrogen by adding H ions
• 8H MnO4-1 ? Mn2 4H2O
• Balance the charge by adding electrons (e-)
• Fe2 ? Fe3 e-
• 5 e- 8H MnO4-1 ? Mn2 4H2O

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To balance a redox reaction in an acidic
solution Example

• Fe2 ? Fe3 e-
• 5 e- 8H MnO4-1 ? Mn2 4H2O
• Multiply the half reactions to equalize the
  number of electrons in each
• 5(Fe2 ? Fe3 e- )
• 5 e- 8H MnO4-1 ? Mn2 4H2O

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To balance a redox reaction in an acidic
solution Example

• 5(Fe2 ? Fe3 e- )
• 5 e- 8H MnO4-1 ? Mn2 4H2O
• Combine the half reactions and cancel anything
  identical (which should always include the
  electrons)
• 5 e- 8H MnO4-1 5Fe2 ? Mn2 4H2O 5Fe3
  5e-

8H MnO4-1 5Fe2 ? Mn2 4H2O 5Fe3
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To balance a redox reaction in an basic solution

• Follow the steps for an acidic solution, then
• To both sides of the reaction, add OH- ions to
  equal the number of H ions remaining
• Form water on the side containing both H and
  OH-, and then cancel water from both sides if
  possible

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To balance a redox reaction in an basic solution
Example

• Balance this redox reaction that takes place in a
  basic solution
• Ag CN-1 O2 ? Ag(CN)2-1 H2O
• Oxidation Ag CN-1 ? Ag(CN)2-1
• Reduction O2 ? H2O

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To balance a redox reaction in an basic solution
Example

• Ag CN-1 ? Ag(CN)2-1 O2 ? H2O
• Ag 2CN-1 ? Ag(CN)2-1 O2 ? H2O
• Ag 2CN-1 ? Ag(CN)2-1 O2 ? 2H2O
• Ag 2CN-1 ? Ag(CN)2-1 4H O2 ? 2H2O

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To balance a redox reaction in an basic solution
Example

• Ag 2CN-1 ? Ag(CN)2-1 4H O2 ? 2H2O
• Ag 2CN-1 ? Ag(CN)2-1 e- 4e- 4H O2 ?
  2H2O
• 4(Ag 2CN-1 ? Ag(CN)2-1 e-) 4e- 4H O2 ?
  2H2O
• 4e- 4H O2 4Ag 8CN-1 ? 4Ag(CN)2-1
  4e- 2H2O

4H O2 4Ag 8CN-1 ? 4Ag(CN)2-1 2H2O
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To balance a redox reaction in an basic solution
Example
4H O2 4Ag 8CN-1 ? 4Ag(CN)2-1 2H2O

• To both sides of the reaction, add OH- ions to
  equal the number of H ions remaining
• 4OH- 4H O2 4Ag 8CN-1 ? 4Ag(CN)2-1 2H2O
  4OH-

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To balance a redox reaction in an basic solution
Example
4OH- 4H O2 4Ag 8CN-1 ? 4Ag(CN)2-1 2H2O
4OH-

• Form water on the side containing both H and
  OH-, and then cancel water from both sides if
  possible
• 4OH- 4H O2 4Ag 8CN-1 ? 4Ag(CN)2-1 2H2O
  4OH-
• 4H2O O2 4Ag 8CN-1 ? 4Ag(CN)2-1 2H2O
  4OH-
• 2H2O O2 4Ag 8CN-1 ? 4Ag(CN)2-1 4OH-

2H2O O2 4Ag 8CN-1 ? 4Ag(CN)2-1 4OH-